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An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass $1\, kg$ moves with a speed of $12 \,m s^{-1}$ and the second part of mass $2\, kg$ moves with $8 \,m s^{-1}$ speed. If the third part files off with $4 \,m s^{-1}$ speed, then its mass is ......... $kg$
$7 $
$17 $
$3$
$5$
Solution
$\begin{array}{l}
\,\,\,\,When\,an\,{\rm{explosion}}\,breaks\,a\,rock,\\
by\,the\,law\,of\,conservation\,of\,momentum,\,\\
initial\,momentum\,is\,zero\,and\,for\,the\\
three\,pieces,\\
Total\,momentum\,of\,the\,two\,pieces\,\\
1\,kg\,and\,2\,kg\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{12}^2} + {{16}^2}} = 20\,kg\,m{s^{ – 1}}.
\end{array}$
$\begin{array}{l}
The\,third\,piece\,has\,the\,same\,momentum\,\\
and\,in\,the\,direction\,opposite\,to\,the\,\\
{\rm{resultant}}\,of\,these\,two\,momenta.\\
\therefore \,Momentum\,of\,the\,third\,pices\\
= 20\,kg\,m{s^{ – 1}}\\
\,\,\,\,\,Velocity\, = 4\,m{s^{ – 1}}\\
\therefore \,Mass\,of\,the\,{3^{rd}}piece\, = \frac{{mv}}{v} = \frac{{20}}{4} = 5kg
\end{array}$
